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Help with hard equation

  1. Nov 22, 2005 #1
    We've got to solve the following equation:
    [itex]i \ Re \ z+Im \ \bar{z}+\pi=|z|+arg(z)[/itex] with [itex]arg(z)\in[0,2\pi)[/itex]
    Please don't solve it for me. Give me a hint first.
    Last edited: Nov 22, 2005
  2. jcsd
  3. Nov 22, 2005 #2


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    First write z= x+ iy= |z|(cos(arg z)+ i sin(arg z))
    of course, [itex]|z|= \sqrt{x^2+ y^2}[/itex] and [itex]arg z= arctan(\frac{y}{x})[/itex].
    Now you can write the condition in terms of x and y.
    Last edited by a moderator: Nov 22, 2005
  4. Nov 22, 2005 #3
    Ok, thanks...

    Did you mean that arg(z)=arctan(y/x)?

    Now it looks like this doesn't it?:
    or [itex](-y-\sqrt{x^2+y^2})+ix=arctan\left(\frac{y}{x}\right)-\pi[/itex]
    Now it seems the number on the right is real, so the number on the left must be real too, so x=0, but then arctan(y/x) is not defined...hm...does it have any consequences?
  5. Nov 22, 2005 #4
    Sure it does, arctan(inf)=pi/2 the answer is then seen to be z=0+i*(pi/4).
  6. Nov 22, 2005 #5
    Ok, thanks guys! :)
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